Question

We roll two fair 6-sided dice. If a fair dice is rolled, the probability of it landing on any number between [1-6] is equi- probable (i.e., 1 6 ). (a) Find the probability that doubles are rolled. (Rolling doubles mean both dices land in the same number, e.g., 3 and 3.) (b) Given that the roll results in a sum of 6 or less, nd the conditional probability that doubles are rolled. (c) Find the probability that the larger of the two die's outcomes is at least 4.

Answer 1

Answer:

a.) $\frac{1}{6}$

b.) $\frac{3}{15}$

c.) $\frac{27}{36}$

Step-by-step explanation:

Given - We roll two fair 6-sided dice. If a fair dice is rolled, the probability

             of it landing on any number between [1-6] is equi- probable

             (i.e., $\frac{1}{6}$).

To find - (a) Find the probability that doubles are rolled.

               (b) Given that the roll results in a sum of 6 or less, and  

                    the conditional probability that doubles are rolled.

               (c) Find the probability that the larger of the two die's outcomes

                    is at least 4.

Proof -

As given , two fair 6-sided dice is rolled.

The sample space is as follows :

S = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

        (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

        (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

        (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

        (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

        (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)  }

Total number of outcomes = 36

a.)

Sample space that double are rolled = { (1, 1), ( 2, 2), ( 3, 3) , ( 4, 4), ( 5, 5), ( 6, 6) }

Total number of possible outcomes = 6

So, Probability = $\frac{6}{36}$ = $\frac{1}{6}$

b.)

Sample space that roll shown sum of 6 or less = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)}

Now,

Total outcomes = 15

Total possible outcome that doubles are rolled = 3 ( i.e (1,1), (2, 2), (3, 3) )

So, Probability = $\frac{3}{15}$

c.)

Sample space that larger of the two die's outcomes is at least 4 = { (1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }

Total possible outcomes = 27

So, Probability = $\frac{27}{36}$